Proof on order of element of cyclic group must divide the order of the group.

Question

I am proving the following problem. I finished proving it but do not know is it sufficient enough.

Problem: Prove that the order of an element in a cyclic group $G$ must divide the order of the group.

My original proof:

Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:

$b^{k}=(a^{m})^{k}$ $\implies$ $e=a^{mk}$ $\implies$$a^n=a^{mk}$ $\implies$ $n=mk$.

Since $m\in \mathbb{Z}$, then $k|n$.

My edited proof:

Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:

$k=\frac{n}{gcd(n,m)} \implies n=(k)gcd(n,m)$

Since $gcd(n,m)\in \mathbb{N}$, then $k|n$.

My question is that I have seen lots of people used Division Algorithm to prove this, but it is only when the element is the generator, here it is any element. Yet, I don't know my prove is sufficient or not. Any suggestion is appreciated.

Answer 1

Your proof is wrong because you state that $a^n=a^{mk}\implies n=mk$. Why do you think that this is true?

However, $a^n=a^{mk}=e\implies n\mid mk$, but this is not enough to deduce that $k\mid n$, which is what you want to prove.

Answer 2

According to your hypothesis that $|G|=n \in {\mathbb N}~$ . Therefore , $G$ is a finite group.

Now for each $a\in G~$ ,$~\langle a\rangle$ forms a subgroup of the group $G$ .Then we see on account of the Lagrange theorem that $|\langle a\rangle |~|~|G|,$ that is ,$|\langle a\rangle |$ is a divisor of $~|G|~.$

Keep in mind that this proof holds for all group $G$ , and hence of course the cyclic one , then our conclusion follows .